Determine vmax and enzyme concentration (Michaelis–Menten)
Michaelis-Menten kinetics is given by
$$V = V_{\max} \frac{S}{K_M + S}$$
Taking the inverse of both sides to linearise the equation,
$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$
Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore,
$$V_\max = \frac{1}{A} = \pu{2.10 mMs^{-1}}$$
$$K_M = B V_\max = \pu{9.39 mM}$$
For the new situation with different $V$ and $S$ and necessarily different $[E_0]$ , $V_{\max}$ would change because $V_{\max} = k_b [E_0]$ but $K_M$ remains constant, thus
$$V_{\max} = V \frac{K_M + S}{S} = \pu{0.101mM s^{-1}} \frac{(\pu{9.39 mM} + \pu{24 mM})}{\pu{24 mM}} = \pu{0.141 mM s^{-1}}$$
You can then calculate $k_b$ by $k_b = V_{\max}/E_0$.