Two lenses of power $$-15 D$$ and $$+5 D$$ are in contact with each other. The focal length of the combination is

For thin lenses kept in contact, their individual powers simply add:
$$P_{\text{eq}} = P_1 + P_2$$

Given lenses have powers $$P_1 = -15 \,\text{D}$$ and $$P_2 = +5 \,\text{D}$$.

Hence
$$P_{\text{eq}} = -15 + 5 = -10 \,\text{D}$$

Power $$P$$ and focal length $$f$$ (in metres) are related by
$$P = \frac{1}{f}\;.$$

Therefore
$$f = \frac{1}{P_{\text{eq}}} = \frac{1}{-10} = -0.10 \,\text{m} = -10 \,\text{cm}\;.$$

The negative sign signifies that the equivalent combination behaves as a diverging lens.

Option B which is: $$-10$$ cm