$$(a,b)$$ We know, $${V_0} = {{Kq} \over R} = Vsurface$$

Now, $${V_i} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {R^2}} \right)\,\,\,\,\,$$ [For $$r < R$$]

At the center of sphere $$r=0.$$ Here

$$V = {3 \over 2}{V_0}$$

Now, $${5 \over 4}{{Kq} \over R} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {r^2}} \right)$$

$${R_2} = {R \over {\sqrt 2 }}$$

$${3 \over 4}{{Kq} \over R} = {{Kq} \over {{R^3}}}$$

$${1 \over 4}{{Kq} \over R} = {{Kq} \over {R{}_4}}$$

$${R_4} = 4R$$

Also, $${R_1} = 0$$ and $${R_2} < \left( {{R_4} - R{}_3} \right)$$