Calculating base resistor value using voltage
It would help a lot if you'd post a link to the datasheet and then cite the line and page for your question.

But in general, if you see a maximum \$V_{BE}\$ value on the datasheet, it is really a \$V_{EBO}\$ value. This just means that if you apply a reverse biasing voltage to the base-emitter junction, that it won't tolerate any more and will instead "break down."

This is about like what a zener diode might do. The BJT isn't supposed to be operated in a reverse-biased base-emitter junction mode. So this is almost  always just a warning about preventing this situation. It means you can tolerate a small amount of reverse-biasing, but not a lot of it. That's all.



As far as forward-biasing goes, the general rule for small signal silicon BJTs is that it takes about \$700\:\text{mV}\$ to produce perhaps \$5\:\text{mA}\$ of collector current. This forward-biased base-emitter voltage increases by about \$60\:\text{mV}\$ if you multiply the collector current by a factor of 10. Simularly, it declines by about \$60\:\text{mV}\$ if you divide the collector current by a factor of 10.

This does NOT mean the base-emitter voltage is fixed, as you can see. It just doesn't vary a lot. So for a lot of reasoned uses, it's close enough for horse-shoes to say \$700\:\text{mV}\$ for the base-emitter junction when operating in active mode. (Which means that the collector voltage is at least \$1\:\text{V}\$ different from the emitter voltage, or more.)

In forward-biased mode, you will not see \$6\:\text{V}\$ across the base-emitter junction (for very long, anyway.) That would pretty much guarantee that the BJT was rapidly destroyed.