Let the final temperature of the mixture be $t$
Heat lost by water at $80^{\circ} C$
$=m s \Delta t $
$=0.1 \times 10^{3} \times s_{\text {water }} \times\left(80^{\circ}-t\right)$
$\left(\because m=V \times d=0.1 \times 10^{3} kg \right)$
Heat gained by water at $60^{\circ} C$
$=0.3 \times 10^{3} \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
According to principle of calorimetry Heat lost = Heat gained
$\therefore \,\,\,\,\,0.1 \times 10^{3} \times s_{\text {water }} \times\left(80^{\circ}-t\right) $
$=0.3 \times 10^{3} \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
or $\,\,\,\,\,\,\left(80^{\circ}-t\right)=3 \times\left(t-60^{\circ}\right)$
or $\,\,\,\,\,\,4 t=260^{\circ}$
or $\,\,\,\,\,\,t=65^{\circ} C$