I don't get the point: why just simply ask for a relation between
$ [T]^B_B \qquad \text{and} \qquad [T^t]^{B^*}_{B^*} \quad \text{?} $
(The one you're asking for would follow from that simpler one, wouldn't it?)
In that case, the relation is simple enough:
$ [T^t]^{B^*}_{B^*} = \left( [T]^B_B \right)^t $
Isn't it?
Remark. I'm assuming that your $T^t$ means the dual of $T$. That is, the linear map $T^t : V^* \longrightarrow V^*$ defined by $T^t\omega = \omega \circ T$. I also assume that your $B^*$ means the dual basis. That is, if $B = \left\{ u_1, \dots , u_n \right\}$ form a basis of $V$, its dual basis is $B^* = \left\{ u_1^*, \dots , u_n^* \right\} $, where the $u_i^* : V \longrightarrow F$ are the linear forms defined by $u_i^*(u_j) = 1 $ if $i =j$ and $0$ otherwise.
If my assumptions are true, I add another piece of standard notation: that $t$ exponent on the right means the transpose of the matrix.
So, in order to prove the stated relation, we can proceed as follows: let's call $A = [T]^B_B$ and display its entries
$ A = \begin{pmatrix} a^1_1 & \dots & a^1_i & \dots & a^1_n \\ \vdots & & \vdots & & \vdots \\ a^j_1 & \dots & a^j_i & \dots & a^j_n \\ \vdots & & \vdots & & \vdots \\ a^n_1 & \dots & a^n_i & \dots & a^n_n \end{pmatrix} $
This means that
$ T(u_i) = a^1_i u_1 + \dots + a^j_iu_j + \dots + a^n_i u_n \ . $
And since we are saying that $[T^t]^{B^*}_{B^*} = A^t$, what we need to show is
$ T^t(u^*_j) = a^j_1 u^*_1 + \dots + a^j_iu^*_i + \dots + a^j_n u^*_n \ , $
for all $j$. Right?
Ok, so now your turn: compute. On one hand, find out
$ T^t(u^*_j) (u_i) $
for every $i,j$. On the other hand, find out
$ ( a^j_1 u^*_1 + \dots + a^j_iu^*_i + \dots + a^j_n u^*_n)(u_i) $
too, for all $i,j$. Compare. Think.