Answer
Oxidizing agent : $H_{2}O$ (l)
Reducing agent : $Na$ (s)
Work Step by Step
First, find oxidation state of each element in the reaction. Oxidation state of Na in element Na is 0. oxidation state of O = -2, H = +1 in $H_{2}O$. In $NaOH$, oxidation state of O= -2, H=+1 and Na=+1. In $H_2$, oxidation state of H=0.
Now, see the change in oxidation state for any atoms. Here, we see that oxidation state of Na is increased (0 in Na to +1 in $NaOH$ ) and oxidation state of H is decreased (+1 in $H_{2}O$ to 0 in $H_2$). That means Na is oxidized (increase in oxidation state) and $H_{2}O$ is reduced. So, oxidizing agent is $H_{2}O$ (because it is reduced but helps Na to oxidize) and reducing agent is $Na$ ( because it is oxidized helps $H_{2}O$ to reduce)
