The question is from Farkas & Kra Riemann surfaces.
I can't make the question self-contained because there're too many definitions, and propositions I need to type. So I'm asking for people who have the book (p.87). But if someone asks for some details and notations via comment, I'll add them.
Proposition. For $g\geq 2$, $q\geq 1$, let $\tau(P)$ be the weight of $P\in M$ with respect to $\mathcal{H}^q(M)$. Let $W_q$ be the Wronskian of a basis for $\mathcal{H}^q(M)$. Set $d = d_q = \dim \mathcal{H}^q(M)$. Then $W_q$ is a (non-trivial) holomorphic $m = m_q$-differential where $\color{\red}{m = (d/2)(2q-1+d)}$.
Proof. We must merely verify that the determinant of the Wronskian $\det[\varphi_1,\ldots,\varphi_d]$ transforms as an $m$-differential under changes of coordinates. Explicitly, let $\{\zeta_1,\ldots,\zeta_d\}$ be a basis for $\mathcal{H}^q(M)$. Let $z$ and $\tilde{z}$ be local coordinates with $\tilde{z} = f(z)$ on the overlap of their respective domains. Assume that $$\zeta_j =\varphi_j(z)dz^q = \tilde{\varphi}_j(\tilde{z})d\tilde{z}^q$$ in terms of the local coordinates $z$ and $\tilde{z}$ (so that $\tilde{\varphi}_j(f(z))f'(z)^q = \varphi_j(z)$). We must show that $$(\det[\varphi_1,\ldots,\varphi_d])dz^m = (\det[\tilde{\varphi}_1,\ldots,\tilde{\varphi}_d])d\tilde{z}^m.\ (\dagger)$$ But it is easy to verify that $$\det[\varphi_1,\ldots,\varphi_d] = \det[(\tilde{\varphi}_1\circ f)(f')^q,\ldots,(\tilde{\varphi}_d\circ f)(f')^q] = (f')^m(\det[\tilde{\varphi}_1,\ldots,\tilde{\varphi}_d]\circ f),$$ which is equivalent to $(\dagger)$. $\square$
I don't understand where the red $m$ comes from. Also, from the last displayed equality, $$\det[(\tilde{\varphi}_1\circ f)(f')^q,\ldots,(\tilde{\varphi}_d\circ f)(f')^q] = (f')^{qd}(\det[\tilde{\varphi}_1,\ldots,\tilde{\varphi}_d]\circ f)$$ from the fact that $\det[f\varphi_1,\ldots,f\varphi_n] = f^n\det[\varphi_1,\ldots,\varphi_n]$ for holomorphic function $f$.
I think the following might imply something...
I was about to delete this post, but it seems many people care about this problem so I post a solution.
In local coordinate $z$, write a basis of $\mathcal{H}^q(M)$ by $\{\varphi_1 dz^q,\ldots,\varphi_d dz^q\}$. Then the Wronskian is by definition $$W_q(z):=\det\begin{pmatrix} \varphi_1(z)dz^q & \cdots & \varphi_d(z)dz^q\\ \vdots & \ddots & \vdots\\ \varphi_1^{(d-1)}(z)dz^{q+d-1} & \cdots & \varphi_d(z)^{(d-1)}dz^{q+d-1}\\ \end{pmatrix}.$$ So, if we focus on the differential part, then the Wronskian is $m$-differential where $$m = q+(q+1)+\cdots+(q+d-1) = (d/2)(2q-1+d).$$