To solve for $D(3\xi)$ given that $\xi \sim B(6,p)$ and $E(2\xi -3)=5$, we proceed as follows:

1. **Find $E(\xi)$:**

Given $E(2\xi -3)=5$, we can express this as:
$$\begin{align*} E(2\xi -3) & = 2E(\xi) - 3 \\ 5 & = 2E(\xi) - 3 \end{align*}$$

Solving for $E(\xi)$, we get:
$$\begin{align*} 2E(\xi) & = 5 + 3 \\ 2E(\xi) & = 8 \\ E(\xi) & = \frac{8}{2} \\ E(\xi) & = 4 \end{align*}$$

2. **Solve for $p$:**

Since $E(\xi) = 6p$, we have:
$$\begin{align*} 6p & = 4 \\ p & = \frac{4}{6} \\ p & = \frac{2}{3} \end{align*}$$

3. **Find $D(\xi)$:**

Knowing that $D(\xi) = 6p(1-p)$, we substitute $p = \frac{2}{3}$:
$$\begin{align*} D(\xi) & = 6 \cdot \frac{2}{3} \cdot \left(1 - \frac{2}{3}\right) \\ & = 6 \cdot \frac{2}{3} \cdot \frac{1}{3} \\ & = 6 \cdot \frac{2}{9} \\ & = \frac{12}{9} \\ & = \frac{4}{3} \end{align*}$$

4. **Calculate $D(3\xi)$:**

Knowing that $D(k\xi) = k^2D(\xi)$ for any constant $k$, we find $D(3\xi)$ by setting $k=3$:
$$\begin{align*} D(3\xi) & = 3^2D(\xi) \\ & = 9 \cdot \frac{4}{3} \\ & = 12 \end{align*}$$

Therefore, the variance of $3\xi$ is $\boxed{12}$.