Note that the events of default for the bonds are independent and identically distributed Bernoulli random variables, hence, the Binomial distribution can be used to calculate the probability of exactly one bond defaulting.

The probability mass function of a binomial random variable, K is given by:

$$P(K = k) = \binom{n}{k} \times p^k \times (1 - p)^{n-k} = \frac{n!}{k!(n - k)!} \times p^k \times (1 - p)^{n-k}$$

Where:
n = the number of bonds in the portfolio
p = the probability of default of each individual bond
k = the number of defaults for which you would like to find the probability.

In our case n = 3, p = 0.10, and k = 1.

So that,

$$P(K = 1) = \frac{3!}{1!(2)!} \times 0.10^1 \times (0.90)^2 = 0.243$$