The accepted answer gets the correct result largely by coincidence. There is not a priori any relationship between the time when the expected number of heads and tails are equal vs. the expected time when the number of heads and tails are equal, and the latter is what is asked for.
To work this out, you can consider a Markov chain for the cumulative number of tails minus the cumulative number of heads. Then this jumps by $+1$ with probability $0.3$, jumps by $-1$ with probability $0.7$, and it starts at $1$. You want the expected time to hit $0$. One way to do this is to consider starting at any $n \geq 0$ and compute the expected times $T(n)$ to hit $0$ starting from all such $n$ together. To do this you apply the total expectation formula to get the recursive relation
$$T(n)=1+0.3 T(n+1)+0.7T(n-1).$$
You can find the general homogeneous solution to this recurrence which is given by $T(n)=c_1 \lambda_1^n + c_2 \lambda_2^n$ where $\lambda_1,\lambda_2=\frac{1 \pm \sqrt{1-4 \cdot 0.3 \cdot 0.7}}{0.6}=\frac{1 \pm 0.4}{0.6}=1,\frac{7}{3}$.
A particular solution can be obtained by the method of undetermined coefficients; you can guess $T(n)=cn$ which yields $cn=1+0.3c(n+1)+0.7c(n-1)=1+cn-0.4c$ yielding $c=5/2$. So the general solution is $T(n)=\frac{5}{2}n + c_1 + c_2 \left ( \frac{7}{3} \right )^n$.
Now clearly $T(0)=0$, which implies that we have $T(n)=\frac{5}{2} n + c \left ( \left ( \frac{7}{3} \right )^n - 1 \right )$ for some $c$.
The subtle question is about what $c$ is. One method for computing it is to consider a modified problem where you also stop the process when you reach a state $N$, and then send $N \to \infty$ (so that this artificial scenario becomes progressively less important). In this case we have $\frac{5}{2} N + c \left ( \left ( \frac{7}{3} \right )^N-1 \right )=0$. You see that then $c \to 0$ as $N \to \infty$, and so $T(n)=\frac{5}{2} n$. Thus the desired quantity is $T(1)=\frac{5}{2}$.