Find total electrons in $B_2^+$

Boron has atomic number $5$.

So, neutral $B_2$ has total electrons: $2 \times 5 = 10$

Therefore, $B_2^+$ has: $10 - 1 = 9 \text{ electrons}$

Write molecular orbital filling for $B_2^+$

For light homonuclear diatomic molecules like $B_2$, the MO order is: $\sigma(1s) < \sigma^*(1s) < \sigma(2s) < \sigma^*(2s) < \pi(2p_x)=\pi(2p_y) < \sigma(2p_z)$

Now fill 9 electrons: $\sigma(1s)^2\, \sigma^*(1s)^2\, \sigma(2s)^2\, \sigma^*(2s)^2\, \pi(2p)^1$

Count unpaired electrons

In $\pi(2p)^1$, there is 1 unpaired electron.

So, $n = 1$

Use spin-only magnetic moment formula

$\mu = \sqrt{n(n+2)}\, \text{BM}$

Substituting $n=1$: $\mu = \sqrt{1(1+2)} = \sqrt{3}\, \text{BM}$

Given: $\sqrt{3} = 1.73$

Hence, $\mu = 1.73\, \text{BM}$

Match with required format

The question asks for: $\underline{\hspace{2cm}} \times 10^{-2} \text{ BM}$

Since $1.73\, \text{BM} = 173 \times 10^{-2} \text{ BM}$

the required integer is: $\boxed{173}$

Comparison with stored answer

Derived answer = $173$

Stored correct answer = $173$

So they agree.