Let $k$ be the smallest integer such that $2^k > 2023^{10}$. Then the game is equivalent to the following: initially is written $1$ and on each move the player adds $1$ or multiples by $2$; winner is who firstly writes a number greater than or equal to $k$. In fact, $k=110$. Indeed, $2023^{10} < 2048^{10} = 2^{110}$, while $2023^{10} > 2^{109}$ is equivalent to $\displaystyle \left(\frac{2023}{2048}\right)^{10} > \frac{1}{2}$, which holds since $\displaystyle \frac{2023}{2048} > \frac{19}{20}$ and $\displaystyle \left(\frac{19}{20}\right)^{10} > \frac{360^5}{400^5} = \frac{9^5}{10^5} = \frac{81 \cdot 729}{100000} > \frac{80 \cdot 700}{100000} = \frac{14}{25} > \frac{1}{2}$. Now we compute from top to bottom the winning and losing positions. (Here we call a position winning if the player whose turn it is now, with this number on the board, has a winning strategy.) The positions $55$, $56$, $\ldots$, $109$ are winning (one doubling is enough), so $28$, $30$, $\ldots$, $54$ are losing and $27$, $29$, $31$, $\ldots$, $53$ are winning, hence $14$, $15$, $\ldots$, $26$ are winning (with a doubling we reach the abovementioned losing positions) and so $3$, $5$, $\ldots$, $13$ are losing and $2$, $4$, $\ldots$, $12$ are winning, therefore $1$ is losing. Since in this game the initial number is $1$, the result of the first move of $A$ is $2$ and therefore $B$ is the winning player.