Answer
$$65 \mathrm{\ s}$
Work Step by Step
With $$P_{\text {cond }} t=L_{V} m=L_{V}(\rho V)=L_{V}(\rho A h),$$
the drop will last a duration of
$$t=\frac{L_{V} \rho A h}{P_{\text {cond }}}=\frac{\left(2.256 \times 10^{6} \mathrm{J} / \mathrm{kg}\right)\left(1000 \mathrm{kg} / \mathrm{m}^{3}\right)\left(4.00 \times 10^{-6} \mathrm{m}^{2}\right)\left(1.50 \times 10^{-3} \mathrm{m}\right)}{0.208 \mathrm{W}}$$$$=65 \mathrm{\ s}$$
