Ringo-tential!

Electric Field at a distance $x$ from the center of a ring having radius $R$ and carrying charge $Q$ is $|E| = \dfrac {qx}{4π\epsilon _0( R^2 + x^2)^{\frac 32}}$

We know by definition that $dV = -Edr$ where $V$ is potential difference and $r$ is distance.

Now, ${\color{#D61F06}{V_1 }} = \displaystyle \int_0^V dV = \displaystyle \int _0^a -\dfrac {qx}{4π\epsilon _0( R^2 + x^2)^{\frac 32}}dx = \dfrac {q}{8π\epsilon _0} \displaystyle \int _0^a -(R^2 + x^2)^{-\frac 32} (2x) dx =\dfrac {q}{8π\epsilon _0} \begin{vmatrix} (R^2 + x^2)^{-\frac 12} \end{vmatrix}_0^a$

$= \dfrac 14 \dfrac {q}{4π\epsilon _0}\begin{pmatrix} \dfrac 1R - \dfrac {1}{\sqrt{ R^2 + a^2}}\end{pmatrix}= {\color{#D61F06}{\dfrac 14 \dfrac {q}{π\epsilon _0R} \begin{pmatrix} 1 - \dfrac {1}{\sqrt {1 + (a/R)^2} }\end{pmatrix}}}$

Similarly for second ring, ${\color{#3D99F6}{V_2 = -\dfrac 14 \dfrac {q}{π\epsilon _0R} \begin{pmatrix} 1 - \dfrac {1}{\sqrt {1 + (a/R)^2} }\end{pmatrix}}}$

Hence, Potential Difference $\Delta V ={\color{#D61F06}{V_1 }}- {\color{#3D99F6}{V_2 }}= \dfrac 12 \dfrac {q}{π\epsilon _0R} \begin{pmatrix} 1 - \dfrac {1}{\sqrt {1 + (a/R)^2}} \end{pmatrix}$

Hence, $a + b = 1 + 2 = \boxed{3}$

Please let me know if my solution is right or wrong