Yes. In the following, which uses (almost) nothing beyond what is already in the problem statement, the calculations involve only simple arithmetic with one-digit numbers (and $10$) and easy estimates involving fractions of two-digit numbers: the stuff of mental arithmetic.
Let $P(k)$ represent the probability of $k$ heads. From the (intuitively obvious) facts that (i) $P(k) \gt 0$ for $0 \le k \le 100$, (ii) $P(k)$ increases from $k=0$ to $k=50$ and then decreases from $k=50$ to $k=100$, and (iii) $P(k) = P(100-k)$, we easily establish the inequalities
$D \gt C, D \gt A, B \gt E.$
I claim that actually $A \gt B$ (i.e., the chance of exactly 50 heads exceeds the chance of 60 or more tails), which establishes $D$ as the answer. To see this, look at the relative probabilities. They all have a common factor of $100!/2^{100}$ which we can ignore, focusing on the binomial coefficients that are left. Now a series of simple estimates establishes
$P(40) / P(50) = \frac{50}{60} \frac{49}{59} \cdots \frac{41}{51} \lt \left(\frac{5}{6}\right)^{10} \lt \frac{1}{1 + 10(1/5)} = \frac{1}{3}.$
(The ratio actually is less than $1/7$.) Moreover,
$P(39) / P(50) = \frac{40}{61} P(40)/P(50) \lt \frac{2}{3} P(40)/P(50).$
Continuing in like vein we see that the chance of $A$ relative to that of $B$, $\left(P(0) + P(1) + \cdots + P(40)\right)/P(50)$, is dominated by a geometric series with starting term $P(40)/P(50)$ and common ratio $2/3$. Therefore its sum is less than $1/3 (1 - 2/3)^{-1} = 1.$ This proves the claim.