[Solution by Dr. Abubacker] Temperature rise $\propto\frac{\text{moles neutralised}\times|\Delta H|}{\text{total volume}}$. HCl + NaOH (strong-strong, $|\Delta H|\approx57$ kJ/mol) with $30$ mmol in $60$ mL gives the largest value. Option (1) has the same amounts but a weak acid (less heat released, part of it spent in ionisation); options (2) and (4) neutralise only $25$ and $20$ mmol in $70$ mL. Answer (3).