Answer
${\bf 0.435}\;\rm \mu A$
Work Step by Step
First of all, we need to find whether the photon energy is greater than the work energy or not, where the work energy is given by $E_0=3.3$ eV.
We know that the energy of the photon is given by
$$E=hf=\dfrac{hc}{\lambda}$$
Plug the known;
$$E =\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{(300\times 10^{-9})}=\bf 6.626\times 10^{-19}\;\rm J=\color{blue}{\bf 4.14}\;\rm eV$$
Now it is obvious that the photon energy is greater than the work energy which means that the electron will be ejected away from the metal plate.
Now we need to find the number of photons that hit the plate in each second.
$$\dfrac{dN_p}{dt}=\dfrac{P}{hf}$$
The author told us that only 12% of the photons eject a photoelectron, this means that the number of electrons that leave the plate in each second is given by
$$\dfrac{N_e}{t}=0.12\left[ \dfrac{P}{hf}\right]$$
where we know that $eN_e/t=q/t=I$,
so we need to multiply both sides by $e$
$$I=0.12e\left[ \dfrac{P}{hf}\right]$$
And $hf=E$ which we found above.
$$I=0.12\;e\left[ \dfrac{ P}{E}\right]$$
Plug the known;
$$I=0.12(1.6\times 10^{-19})\left[ \dfrac{(15\times 10^{-6})}{( 6.626\times 10^{-19})}\right]$$
$$I=\color{red}{\bf 0.435}\;\rm \mu A$$
