Answer: 573 J

Here, $H_2SO_4$ is the limiting reactant.

Since, $N_1V_1 = N_2V_2$

$0.2 \times 50 = 1 \times V_2$

$V_2 = \frac{2}{10} \times 50$ = 10 mL

Thus, 50 mL of 0.2 N $H_2SO_4$ will exactly neutralise with 10 mL of 1 N KOH.

Heat produced on neutralisation of 1000 mL of 1 N KOH = 57.3 kJ

$\therefore$ Heat produced on neutralisation of 10 mL of 1 N KOH = $\frac{57.3}{1000}\times 10 \times 1000 $ J = 573 J