The standard way is to find the CDF of $\hat{\theta}_n$ first, which is
\begin{align*}
F(x) := P(\max(X_1, \ldots, X_n) \leq x) = \prod_{i = 1}^n \frac{x}{\theta} =
\frac{x^n}{\theta^n}, \; 0 \leq x \leq \theta.
\end{align*}
This implies the PDF of $\hat{\theta}_n$ is $f(x) = nx^{n - 1}/\theta^n$ for $0 \leq x \leq \theta$ and $0$ otherwise, whence
\begin{align*}
E[(\hat{\theta}_n - \theta)^2] = \int_0^\theta(t - \theta)^2f(t)dt =
\frac{n}{\theta^n}\int_0^\theta (t - \theta)^2t^{n - 1}dt. \tag{1}
\end{align*}
To evaluate the integral $(1)$, apply the variable substitution $u = t/\theta$, which transfers $(1)$ to
\begin{align*}
n\int_0^1\theta^2(1 - u)^2 u^{n - 1} du = \theta^2nB(3, n), \tag{2}
\end{align*}
where $B(a, b)$ is the Beta function. Since $B(a, b) = (a - 1)!(b - 1)!/(a + b - 1)!$ when $a, b$ are positive integers, the right-hand side of $(2)$ equals to
\begin{align}
\theta^2 \frac{2n(n - 1)! }{(n + 2)!} = \frac{2\theta^2}{(n + 2)(n + 1)},
\end{align}
which converges to $0$ as $n \to \infty$. This shows that $\hat{\theta}_n \to \theta$ in quadratic mean.
The above method is straightforward. However, it involves pretty heavy calculations (you may circumvent summoning the Beta function by expanding $(1 - u)^2$ then applying linearity of the integral, which requires less machinery). This motivates me to consider the following alternative solution.
First, show that $\hat{\theta}_n$ converges to $\theta$ in probability. I will leave this to you as an exercise. It is similar to finding the CDF of $\hat{\theta}_n$.
Next, for any $\epsilon \in (0, \theta)$, we have
\begin{align*}
& E[(\hat{\theta}_n - \theta)^2] =
E\left[(\hat{\theta}_n - \theta)^2I_{[|\hat{\theta}_n - \theta| \leq \epsilon]}(\omega)\right] +
E\left[(\hat{\theta}_n - \theta)^2I_{[|\hat{\theta}_n - \theta| > \epsilon]}(\omega)\right] \\
\leq & \epsilon^2 + 4\theta^2P[|\hat{\theta}_n - \theta| > \epsilon]. \tag{3}
\end{align*}
Since $\hat{\theta}_n$ converges to $\theta$ in probability, the second term in $(3)$ converges to $0$ as $n \to \infty$. Since $\epsilon$ is arbitrary, this implies that $E[(\hat{\theta}_n - \theta)^2] \to 0$ as $n \to \infty$, completing the proof.
