Answer
$2.13\;\rm kV$
Work Step by Step
We know that the potential at any point is the superposition of the potentials due to all charges [as the author mentioned as a hint].
Let's call the big charged sphere as 1 and the other one as 2.
So, the potential at point a is given by
$$V_a=V_1+V_2$$
From the given figure,
$$V_a=\dfrac{kQ_1}{R_1}+\dfrac{kQ_2}{1-R_1}$$
$$V_a=k\left[\dfrac{ Q_1}{R_1}+\dfrac{ Q_2}{1-R_1}\right]\tag 1$$
The potential at point b is given by
$$V_b=V_1+V_2$$
From the given figure,
$$V_b=\dfrac{kQ_1}{1-R_2}+\dfrac{kQ_2}{R_2}$$
$$V_b=k\left[\dfrac{ Q_1}{1-R_2}+\dfrac{Q_2}{R_2}\right]\tag 2$$
Thus, the potential difference between points a and b is given by
$$\Delta V=V_b-V_a$$
Plug from (1) and (2),
$$\Delta V=k\left[\dfrac{ Q_1}{1-R_2}+\dfrac{Q_2}{R_2}\right]-k\left[\dfrac{ Q_1}{R_1}+\dfrac{ Q_2}{1-R_1}\right]$$
Plugging the known;
$$\Delta V=(9\times 10^9)\left[\dfrac{ 100\times 10^{-9}}{1-0.05}+\dfrac{25\times 10^{-9}}{0.05}\right]-\\
(9\times 10^9)\left[\dfrac{ 100\times 10^{-9}}{0.30}+\dfrac{ 25\times 10^{-9}}{1-0.30}\right]$$
$$\Delta V=\color{red}{\bf 2126}\;\rm V$$
It is obvious now that $V_b\gt V_a$ since the final result is positive.