Answer
the rate should be $\frac{375}{128}\pi+2$.
Work Step by Step
let radius be r, height be h, slant height be l, amount poured in be x, time be t.
$\ 16/5=h/r $
$\ r=5/16h $
as rate of oozing out is proportional to area in contact,
rate of oozing out = $\ a \pi r l$
by Pythagoras theorem, $\ l= \sqrt (r^2+h^2) $
change in amount of liquid = rate of pouring in - rate of oozing out
$\ \frac{d}{dt} (1/3 \pi r^2 h) = \frac{d}{dt} (x) - a \pi r l $
by substituting $\ r $ and $\ l $ and differentiating, we get
$\ \frac{75 \pi}{768} h^2 \frac{dh}{dt} = \frac{dx}{dt} - \frac{5 \sqrt 281}{256} a \pi h^2 $
substitute $\ \frac{dh}{dt} $ = -0.3, $\ \frac{dx}{dt} $ = 2, $\ h $=10
, we get $\ a = \frac {\frac{-375}{128} \pi -2} {\frac{-500 \sqrt 281}{256} \pi} $
substituting $\ a = \frac {\frac{-375}{128} \pi -2} {\frac{-500 \sqrt 281}{256} \pi} $ , $\ h $=10, $\ \frac{dh}{dt} $ = 0,
we get $\ \frac{dx}{dt} $ = $\frac{375}{128}\pi+2$
therefore the rate should be $\frac{375}{128}\pi+2$.
