1. Use the condition for minima in single-slit diffraction

For a slit of width $a$, the minima occur at

$a\sin\theta = m\lambda, \qquad m=1,2,3,\dots$

The second minima from the central maximum corresponds to

$m=2$

So,

$a\sin\theta = 2\lambda$

2. Convert the given slit width into SI units

Given:

$a = 22.0\times 10^{-5}\,\text{cm}$

Since $1\,\text{cm}=10^{-2}\,\text{m}$,

$a = 22.0\times 10^{-5}\times 10^{-2}\,\text{m} = 22.0\times 10^{-7}\,\text{m} = 2.2\times 10^{-6}\,\text{m}$

Wavelength:

$\lambda = 550\,\text{nm} = 550\times 10^{-9}\,\text{m} = 5.5\times 10^{-7}\,\text{m}$

3. Substitute into the minima condition

$\sin\theta = \frac{2\lambda}{a}$

$\sin\theta = \frac{2(5.5\times 10^{-7})}{2.2\times 10^{-6}}$

$\sin\theta = \frac{11\times 10^{-7}}{2.2\times 10^{-6}} = 0.5$

Thus,

$\theta = \sin^{-1}(0.5) = \frac{\pi}{6}$

4. Match with the options

$\frac{\pi}{6}$ corresponds to Option C.

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.