Answer
The potential at point $a$ is $5~V$
The potential at point $b$ is $-2~V$
Work Step by Step
We can assume that the current flows clockwise. We can use the loop rule to find the current in the circuit:
$-(1~\Omega)~I+15~V - (4~\Omega)~I -5~V = 0$
$(5~\Omega)~I = 10~V$
$I = \frac{10~V}{5~\Omega}$
$I = 2~A$
We can find the potential difference across the $4~\Omega$ resistor:
$\Delta V = I~R = (2~A)(4~\Omega) = 8~V$
From ground to point a, there is a potential increase of 5 V. Therefore, the potential at point a is 5 V.
From point a across the $4~\Omega$ resistor, the potential increases by 8 V. Then the potential decreases 15 V before reaching point b. Therefore, the potential at point b is $5~V +8~V-15~V$ which is $-2~V$
