The problem is given by:
$$ \begin{align*} \arg \min_{\boldsymbol{x}} \quad & \frac{1}{2} {\left\| \boldsymbol{X} - \boldsymbol{B} \right\|}_{F}^{2}, \; \boldsymbol{B} \in \mathbb{R}^{n \times n} \\ \text{subject to} \quad & \begin{aligned} \operatorname{Tr} \left( \boldsymbol{X} \right) & = 1 \\ \end{aligned} \end{align*} $$
The Lagrangian is given by:
$$ L \left( \boldsymbol{X}, \mu \right) = \frac{1}{2} {\left\| \boldsymbol{X} - \boldsymbol{B} \right\|}_{F}^{2} + \mu \left( \operatorname{Tr} \left( \boldsymbol{X} \right) - 1 \right) $$
The Lagrangian derivative with respect to $ \boldsymbol{X} $:
$$ \nabla_{\boldsymbol{X}} L \left( \boldsymbol{X} \right) = \boldsymbol{X} - \boldsymbol{B} + \mu \boldsymbol{I} $$
Taking the trace of each element and equating to zero will yield:
$$ \operatorname{Tr} \left( \boldsymbol{X} \right) - \operatorname{Tr} \left( \boldsymbol{B} \right) + \mu \operatorname{Tr} \left( \boldsymbol{I} \right) = 1 - \operatorname{Tr} \left( \boldsymbol{B} \right) +\mu n \implies \mu = \frac{ \operatorname{Tr} \left( \boldsymbol{B} \right) - 1 }{n} $$
Since we can also extract the optimal $\boldsymbol{X}$ the end result is given by:
$$ \boldsymbol{X} = \boldsymbol{B} - \frac{\operatorname{Tr} \left( \boldsymbol{B} \right) - 1}{n} \boldsymbol{I} $$