1. Given probability distribution

The random variable $X$ takes values $0,1,2,3$ with probabilities

$$P(X=0)=\frac{1-d}{4},\quad P(X=1)=\frac{1+2d}{4},\quad P(X=2)=\frac{1-4d}{4},\quad P(X=3)=\frac{1+3d}{4}.$$

2. Use validity of probabilities

For a probability distribution, each probability must be non-negative.

So,

$$\frac{1-d}{4} \ge 0 \Rightarrow 1-d\ge 0 \Rightarrow d\le 1$$ $$\frac{1+2d}{4} \ge 0 \Rightarrow 1+2d\ge 0 \Rightarrow d\ge -\frac12$$ $$\frac{1-4d}{4} \ge 0 \Rightarrow 1-4d\ge 0 \Rightarrow d\le \frac14$$ $$\frac{1+3d}{4} \ge 0 \Rightarrow 1+3d\ge 0 \Rightarrow d\ge -\frac13$$

Combining all conditions,

$$-\frac13 \le d \le \frac14.$$

Hence, the minimum possible value of $d$ is

$$d=-\frac13.$$

3. Find the mean $E(X)$

$$E(X)=\sum xP(X=x).$$

So,

$$E(X)=0\cdot \frac{1-d}{4}+1\cdot \frac{1+2d}{4}+2\cdot \frac{1-4d}{4}+3\cdot \frac{1+3d}{4}.$$

Simplify:

$$E(X)=\frac{1+2d+2-8d+3+9d}{4}$$ $$E(X)=\frac{6+3d}{4}.$$

Now substitute $d=-\frac13$:

E(X)=\frac{6+3\left(-\frac13\right)}{4}=rac{6-1}{4}=rac54.

4. Compute sixty times the mean

$$60\times E(X)=60\times \frac54=15\times 5=75.$$

Therefore,

$$\boxed{75}$$