Answer : 1250 s

Using the expression $m=(Q/F)(M/|v_c|)$

We get

$(80\times 5\times 10^{-3}cm^3)(10.5 gcm^{-3})=\large\frac{(3A)t}{(96500Cmol^{-1})}\big(\large\frac{108gmol^{-1}}{1}\big)$

This gives $t=\large\frac{(80\times 5\times 10^{-3})(10.5)(96500)}{3(108)}=$1251 s