$q_0$ is start state

**$\mathbf{\delta(q_0,0,Z_0) = (q_0,Z_0)}$ **

[ Do Nothing operation, just read any no of $0$'s but do not keep in stack (any no of $0$'s because on reading $0$'s it remains on same state $q_0$) ]

**$\mathbf{\delta(q_0,1,Z_0) = (q_0,XZ_0)}$ ** [ Read first $1$ and keep one $X$ in stack ]

**$\mathbf{\delta(q_0,1,X) = (q_0,XX)}$ ** [Read any no of $1$'s and keep one $X$ for each $1$ in stack ]

**$\mathbf{\delta(q_0,0,X) = (q_1,X)}$ **

[ Read single $0$ and do nothing in stack, state changed from $q_0$ to $q_1$ ]

**$\mathbf{\delta(q_1,1,X) = (q_1,\epsilon)}$ **

[ Pop out one $X$ from stack on reading each $1$ on state $q_1$ (matching each $1$ with the $1$ read before single $0$ ) ]

**$\mathbf{\delta(q_0,\epsilon,Z_0) = (q_0,\epsilon)}$ **

[stack is empty , inputs are accepted here ,that is , $\epsilon$ or any of $0$'s (we read earlier with Do Nothing operation) ]

**$\mathbf{L = \{ 0^m , m \geq 0 \}}$**

No input accept after reaching on $q_1$ because stack will remain with $Z_0$, stack initial symbol

Note : if we add one more transition **$\mathbf{\delta(q_1,\epsilon,Z_0) = (q_1,\epsilon)}$ ****,** then $L$ will be $\{ 0^m \bigcup 0^i1^j01^j ,m,i,j \geq 0 \}$