Answer
$\approx 2.7\;\rm h$
Work Step by Step
We know that the heat needed to warm up the alligator is given by
$$Q=mc\Delta T\tag 1$$
And to find the time it would take to warm up from 23$^\circ$C to 30$^\circ$C, we need to find the power energy absorbed by the alligator which is given by
$$P=IA\tag 2$$
where $I$ is the sun's intensity and $A$ is the alligator's surface area.
Recalling that
$$P=\dfrac{Q}{t}$$
where $t$ is the time
Plugging into (2);
$$\dfrac{Q}{t}=IA$$
Hence,
$$t=\dfrac{Q}{IA}$$
Plugging from (1);
$$t=\dfrac{mc\Delta T}{IA}$$
Plugging the known;
$$t=\dfrac{(350)(3400)(30-23)}{(500)(2.9\times 0.6)}$$
$$t=\color{red}{\bf 9.6\times 10^3}\;\rm s\approx\bf 2.7\;\rm h$$
