First EM problem that I solved by myself

We Use Energy Conservation ! (take reference for gravitational Potential energy at disc)

$\displaystyle{ U }_{ f,grav. }+{ K }_{ f }+{ U }_{ f,Elect }\quad ={ \quad U }_{ i,grav. }+{ K }_{ i }+{ U }_{ i,Elect }$ .

$\displaystyle0+0+q{ V }_{ f }\quad =\quad mgH+0+q{ V }_{ i }\quad \\ q(\cfrac { \sigma }{ 2\epsilon } R)\quad =\quad mgH+q(\cfrac { \sigma }{ 2\epsilon } (\sqrt { { H }^{ 2 }+{ R }^{ 2 } } -H))$ .

after solving this this simple equation by substituting the given value of specific Charge ratio we should get :

$\boxed { H=\cfrac { 2R }{ 3 } }$ . Answer

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Note :Here I used Directly General Standard expression for electric Potential at hight 'x' is given by : $\\ \boxed { V(x)\quad =\cfrac { \sigma }{ 2\epsilon } (\sqrt { { x }^{ 2 }+{ R }^{ 2 } } -x) }$ .

Dear Aman, you gave a wrong expression for the mass (m) versus charge (q) relation m/q, if you really want the readers to get your answer (7).