Answer: 15

Small Hint:

Use $\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}$

Big Hint:

Combine the four inverse-cotangent angles two at a time from left to right

Solution:

Let $\alpha=\cot^{-1}3+\cot^{-1}7.$ The cotangent addition formula gives $$\begin{aligned} \cot\alpha&=\frac{3\cdot7-1}{3+7}\\ &=2. \end{aligned}$$ Combining the next angle gives $$\frac{2\cdot13-1}{2+13}=\frac53,$$ and combining the last gives $$\frac{(\frac{5}{3})\cdot21-1}{\frac{5}{3}+21}=\frac32.$$ Multiplying by $10$ yields $15.$