A Wheatstone bridge is initially at room temperature and all arms...

Initially, the Wheatstone bridge is balanced with $R_1 = R_2 = R_3 = R_4 = R$.
Since the bridge is balanced, the potential difference across the bridge is zero: $V_a - V_b = 0$ V.
When $R_3$ is heated and its resistance increases by 10%: $R_3' = 1.1R$
The voltage at node $a$ (between $R_1$ and $R_2$): $V_a = 40 \times \frac{R_2}{R_1 + R_2} = 40 \times \frac{R}{R + R} = 40 \times \frac{1}{2} = 20$ V
The voltage at node $b$ (between $R_3'$ and $R_4$): $V_b = 40 \times \frac{R_4}{R_3' + R_4} = 40 \times \frac{R}{1.1R + R} = 40 \times \frac{1}{2.1} \approx 19.05$ V
Therefore: $V_a - V_b = 20 - 19.05 = 0.95$ V