Answer:

$\dim(K) = \dim(X)$, because $f = \frac{X} {X + K}$.
f must be dimensionless, because $f = \frac{X} {X + K}$.
$\dim(V_b) = \dim(V) = \mbox{length}^3$, because V(0) = V_b.
$\dim(V_m) = \dim(V) = \mbox{length}^3$, because $V(\infty) = f^3 V_m$, and f is dimensionless.
$\dim(\dot{r}_B) = \mbox{time}^{-1}$, because $t \dot{r}_B$ must be dimensionless; it occurs as an argument of a transendental function.
$\dim(\dot{k}_M) = \dim(\dot{r}_B) = \mbox{time}^{-1}$, because $\dot{r}_B = (3/\dot{k}_M + 3 f V_m^{1/3}/\dot{v})^{-1}$.
$\dim(\dot{v}) = \mbox{length}/\mbox{time}$, because $\dim((f V_m^{1/3}/\dot{v})^{-1}) = \dim(\dot{r}_B) = \mbox{time}^{-1}$.
g must be dimensionless because $\dim(\frac{\dot{v}} {g \dot{k}_M}) = \dim(V_m^{1/3}) = \mbox{length}$; $\dim(\dot{v})$ and $\dim(\dot{k}_M)$ are known.