The rolling starts now...

Hope the Pic Helps!!

$Let\quad the\quad mass\quad of\quad the\quad wheel\quad me\quad 'M'.\quad It's\quad radius\quad is\quad 'R'\\ and\quad it's\quad angular\quad velocity\quad is\quad '\omega '.\quad Let\quad it\quad rotate\quad in\quad the\quad \\ clockwise\quad direction.\quad Since\quad the\quad friction(f)\quad decelerates\\ the\quad rotation,\quad so,\quad it's\quad direction\quad is\quad towards\quad right,\quad and\\ thus\quad the\quad body\quad accelerates\quad towards\quad right.\quad Let's\quad it\\ angular\quad acceleration\quad is\quad '\alpha '\quad and\quad it's\quad linear\\ acceleration\quad is\quad 'a'.\\ \\ Using\quad the\quad info.\quad for\quad translational\quad motion,\quad f=Ma---1.\\ Using\quad the\quad info.\quad for\quad rotational\quad motion,\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \tau =I\alpha \\ where\quad \tau \quad is\quad the\quad applied\quad torque\quad and\quad I\quad is\quad the\quad moment\\ of\quad inertia.\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad R(f)\quad =\quad \frac { M{ R }^{ 2 } }{ 2 } \alpha \\ Using\quad 1.\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad R(Ma)\quad =\quad \frac { M{ R }^{ 2 } }{ 2 } \alpha \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a=\frac { R }{ 2 } \alpha \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \alpha \quad =\quad \frac { 2a }{ R } \\ But\quad friction\quad f\quad =\quad \mu Mg\\ Therefore,\quad a\quad =\quad \mu g\\ So,\quad \alpha \quad =\quad \frac { 2\mu g }{ R } \\ \\ \quad \quad \quad \quad \quad \quad \quad \quad$

$Now,\quad writing\quad the\quad equation\quad for\quad translational\quad motion,\\ the\quad wheel's\quad velocity\quad after\quad time\quad 't'\quad =\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { v }_{ f }={ v }_{ i }+at\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { v }_{ f }=0+\mu gt\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { v }_{ f }=\mu gt\quad ------2.\\ \\ For\quad rotational\quad motion,\quad the\quad wheels\quad angular\quad velovity,\\ after\quad time\quad 't'\quad =\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { \omega }_{ f }={ \omega }_{ i }-\alpha t\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { \omega }_{ f }=\omega -\frac { 2\mu gt }{ R } \quad ---3.\\ But,\quad for\quad pure\quad rolling,\quad { v }_{ f }=R{ \omega }_{ f }\\ Using\quad 2.\quad \& \quad 3.\\ \mu gt\quad =\quad R(\omega -\frac { 2\mu gt }{ R } )\\ \mu gt\quad =\quad R\omega \quad -\quad 2\mu gt\\ 3\mu gt\quad =\quad R\omega \\ t\quad =\quad \frac { 1 }{ 3 } \frac { R\omega }{ \mu g } \\ Comparing\quad with\quad the\quad given\quad equation,\quad a=1,\quad b=3\\ a+b=4$

Simply apply Conservation of Angular Momentum About Icr (point of contact in this case).And we get w'=w/3. Now wR/3= $\mu$ gt

So t= $\frac { 1 }{ 3 } \frac { R\omega }{ \mu g }$ so 1+3=4