Answer
$\approx 24\;\rm W$
Work Step by Step
We know that the power needed is given by
$$P=F_{x,gardener}v $$
$$P=F_{gardener}\;v\cos37^\circ\tag 1 $$
So, we have to find the magnitude of the force applied by the gardener. And to do so, we need to draw the force diagram of the lawnmower.
We can apply Newton's second law. And as we see that the lawnmower is moving at a constant velocity which means that the net force exerted on it is zero.
Thus,
$$\sum F_x=F_{gardener}\cos37^\circ-F_k=ma_x=m(0)=0$$
Hence,
$$F_{gardener}\cos37^\circ=F_k$$
Recall that $F_k=\mu_kF_n$,
$$F_{gardener}\cos37^\circ=\mu_kF_n \tag 2$$
And,
$$\sum F_y=F_n-F_{gardener}\sin37^\circ-mg=ma_y=m(0)=0$$
Hence,
$$F_n=F_{gardener}\sin37^\circ+mg$$
Plugging into (2);
$$F_{gardener}\cos37^\circ=\mu_k \left(F_{gardener}\sin37^\circ+mg\right)$$
Solving for $F_{gardener}$;
$$F_{gardener}\cos37^\circ=\mu_k F_{gardener}\sin37^\circ+\mu_k mg $$
$$F_{gardener}\cos37^\circ-\mu_k F_{gardener}\sin37^\circ=\mu_k mg $$
$$F_{gardener}\left[\cos37^\circ-\mu_k \sin37^\circ\right]=\mu_k mg $$
$$F_{gardener} =\dfrac{\mu_k mg}{\cos37^\circ-\mu_k \sin37^\circ} $$
Plugging into (1);
$$P=\left(\dfrac{\mu_k mg}{\cos37^\circ-\mu_k \sin37^\circ} \right)v \cos37^\circ$$
Plugging the known;
$$P=\left(\dfrac{0.15\times 12\times 9.8}{\cos37^\circ-0.15\sin37^\circ} \right) \times 1.2\cos37^\circ $$
Therefore,
$$P=\color{red}{\bf 23.86 }\;\rm W$$
