Answer
a) $V_b\gt V_a=V_c$
b) $V_a\gt V_b\gt V_c$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric potential is given by
$$V=\dfrac{kq}{r}$$
The two charges are positive, so
$$V_{a}=V_1+V_2 $$
$$V_a= \dfrac{kq}{r_1}+ \dfrac{kq}{r_2}$$
$$V_a= \dfrac{kq}{r_1}+ \dfrac{kq}{3r_1}=\dfrac{3kq+kq}{3r_1}$$
Let's assume that $r_1=1$ unit.
$$V_a= \dfrac{4kq }{3r_1}=\boxed{1.3kq}$$
By the same approach,
$$V_c= \dfrac{4kq }{3r_1}=\boxed{1.3kq}$$
And
$$V_b=\dfrac{kq}{1}+ \dfrac{kq}{1}=\boxed{2kq}$$
Therefore,
$$\boxed{V_b\gt V_a=V_c}$$
$$\color{blue}{\bf [b]}$$
$$V_{a}=V_1+V_2 $$
$$V_a= \dfrac{kq}{r_1}+ \dfrac{-kq}{r_2}$$
$$V_a= \dfrac{kq}{r_1}+ \dfrac{-kq}{3r_1}=\dfrac{3kq-kq}{3r_1}$$
Let's assume that $r_1=1$ unit.
$$V_a= \dfrac{2kq }{3r_1}=\boxed{0.67kq}$$
$$V_a= \dfrac{kq}{3r_1}+ \dfrac{-kq}{r_1}=\dfrac{kq-3kq}{3r_1}$$
$$V_c= \dfrac{4kq }{3r_1}=\boxed{-0.67kq}$$
And
$$V_b=\dfrac{kq}{r_1}- \dfrac{kq}{r_1}=\boxed{0}$$
Therefore,
$$\boxed{V_a\gt V_b\gt V_c}$$
