= Solution
Adjacent slit contributions have <phase difference> $\delta=2\pi d\sin\theta/\lambda$. The six-slit <amplitude> is a <geometric sum> $\sum_{j=0}^{5}e^{ij\delta}=e^{5i\delta/2}\sin(3\delta)/\sin(\delta/2)$. The first zero away from the central maximum occurs at $\delta=2\pi/6$, where the denominator is nonzero. Therefore
$$
\boxed{\theta_1=\arcsin\frac{\lambda}{6d}=\arcsin\frac{550\times10^{-9}}{6\times100\times10^{-6}}\simeq9.17\times10^{-4}\,\mathrm{rad}=0.0525^\circ.}
$$
There is a matching zero at the negative <angle>. This is the <first minimum of a finite slit array>, rather than the much larger principal-order <angle> or a single-slit envelope minimum. The slits are stipulated sufficiently narrow for that envelope not to set the first zero.
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