Hi Vinoth, here are my thoughts, hopefully they're correct and what you're after:

You can think of $\mathfrak{\tilde{g}}$ as the set of pairs

$$\{(X,L)\in \mathfrak{g}\times \mathbb{P}^{1}\;|\; X(L)\subset L\}$$

(identify $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$ via the Killing form). Thus, the fibre over $L\in \mathbb{P}^{1}$ is the the set of $X\in \mathfrak{sl}_{2}$ that have $L$ as an 'eigenline'. Since the projection to $G/B$ is $G$-equivariant it suffices to consider a particular line $L_{0}$; take $L_{0}=span\{e_{1}\}$, to see that the fibre is the standard Borel $\mathfrak{b}$, so that the fibre over $g\cdot L_{0}$ is $g\cdot\mathfrak{b}$ .

Hi Vinoth, here are my thoughts, hopefully they're correct and what you're after:

You can think of $\mathfrak{\tilde{g}}$ as the set of pairs

$$\{(X,L)\in \mathfrak{g}\times \mathbb{P}^{1}\;|\; X(L)\subset L\}$$

(identify $\mathfrak{g}$ and $\mathfrak{g}^{\ast}$ via the Killing form). Thus, the fibre over $L\in \mathbb{P}^{1}$ is the the set of $X\in \mathfrak{sl}_{2}$ that have $L$ as an 'eigenline'. Since the projection to $G/B$ is $G$-equivariant it suffices to consider a particular line $L_{0}$; take $L_{0}=span\{e_{1}\}$, to see that the fibre is the standard Borel $\mathfrak{b}$, so that the fibre over $g\cdot L_{0}$ is $g\cdot\mathfrak{b}$.

EDIT: providing the answer in terms of the dual $\mathfrak{g}^{\ast}$, as the question states, we see that the fibre is the span of the dual basis (via the Killing form) of the standard basis $\{F,H\}$ of the lower triangular Borel.