Consider $\mathrm{P}$ be a point at distance $\mathrm{r}$ from $\mathrm{O}$ and $\mathrm{OP}$ makes an angle $\theta$ with z-axis. Component of $\mathrm{M}$ along $\mathrm{OP}=\mathrm{M} \cos \theta$. So, the magnetic field induction at $\mathbf{P}$ due to dipole of moment $\operatorname{Mcos} \theta$ is
$$
\mathrm{B}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M} \cos \theta}{\mathrm{r}^3} \hat{\mathrm{r}}
$$
According to the diagram, $\mathrm{r}$ is the radius of sphere with centre at $\mathrm{O}$ lying in $\mathrm{yz}$-plane. Take an elementary area $\mathrm{dS}$ of the surface at $P$. So,
$$
\mathrm{dS}=\mathrm{r}(\mathrm{r} \sin \theta \mathrm{d} \theta) \hat{\mathrm{r}}=\mathrm{r}^2 \sin \theta \mathrm{d} \theta \hat{\mathrm{r}}
$$
$$
\left[\therefore \mathrm{d} \theta=\frac{\mathrm{dS}}{\mathrm{r}^2} \text { or } \mathrm{dS}=\mathrm{r}^2 \mathrm{~d} \theta\right]
$$
$$
\begin{aligned}
&\oint \mathrm{B} \cdot \mathrm{dS}=\oint \frac{\mu_0}{4 \pi} \frac{2 \mathrm{M} \cos \theta}{\mathrm{r}^3} \hat{\mathrm{r}}\left(\mathrm{r}^2 \sin \theta \mathrm{d} \theta \hat{\mathrm{r}}\right) \\
&=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}} \int_0^{2 \pi} 2 \sin \theta \cdot \cos \theta \mathrm{d} \theta=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}} \int_0^{2 \pi} \sin 2 \theta \mathrm{d} \theta \\
&=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{\mathrm{r}}\left(\frac{-\cos 2 \theta}{2}\right)_0^{2 \pi} \\
&=-\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{2 \mathrm{r}}[\cos 4 \pi-\cos 0]=\frac{\mu_0}{4 \pi} \frac{\mathrm{M}}{2 \mathrm{r}}[1-1]=0
\end{aligned}
$$