The correct answer is $\boxed{\text{B) 0.1937}}$.
To calculate the probability of exactly 2 defective items in a sample of 10 items, we can use the binomial distribution. The binomial distribution is a probability distribution that describes the number of successes in a sequence of independent experiments each of which yields success with probability $p$. In this case, the experiment is choosing an item from the lot, and the success is choosing a defective item. The probability of success is $p=0.1$, since 10% of the items in the lot are defective. We want to know the probability of getting 2 successes in 10 trials, so $n=10$.
The probability of exactly 2 defective items is given by the following formula:
$$P(X=2)=\binom{n}{2}p^2(1-p)^{n-2}$$
where $\binom{n}{2}$ is the binomial coefficient, which is the number of ways to choose 2 items from a set of 10 items. In this case, $\binom{10}{2}=45$.
Plugging in $p=0.1$ and $n=10$, we get:
$$P(X=2)=\binom{10}{2}(0.1)^2(0.9)^{10-2}=0.1937$$
Therefore, the probability of exactly 2 defective items in a sample of 10 items is $\boxed{\text{B) 0.1937}}$.
The other options are incorrect because they do not represent the probability of exactly 2 defective items in a sample of 10 items.